Question: The equation of a circle $C$ is $x^2+y^2+10x-8y+25 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+10x) + (y^2-8y) = -25$ $(x^2+10x+25) + (y^2-8y+16) = -25 + 25 + 16$ $(x+5)^{2} + (y-4)^{2} = 16 = 4^2$ Thus, $(h, k) = (-5, 4)$ and $r = 4$.